A particle of charge `-16xx10^(-18) coulomb` moving with velocity `10 ms^(-1)` along the ` x- axis `, and an electric field of magnitude `(10^(4))//(m)` is along the negative ` z- ais`. If the charged particle continues moving along the ` x- axis` , the magnitude of `B` is
A
`(10^(3) Wb)//(m^(2))`
B
`(10^(5) Wb)//(m^(2))`
C
`(10^(16) Wb)//(m^(2))`
D
`(10^(-3) Wb)//(m^(2))`
Text Solution
Verified by Experts
The correct Answer is:
A
The situation is shown in the figure . `F_(E) = Force due to electric field ` `F_(B) = Force due to magnetic field ` It is given that the charged particle remins moving along `X-axis`( i.e. undeviated). Therefore `F_(B) = F_(E)` rArr qvB = qE rArr B = (E)/(v) = (10^(4))/(10) = 10^(3) weber//m^(2)`
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